Question

find the zeros of the quadratic polynomial of x2-4 and verify the relationship between zeroes and coefficientsata​

Answer 1

Answer:

Step-by-step explanation:

Let p(x)=x2−4=x2−22=(x−2)(x+2)

∴ p(x)=0

⇒ (x−2)(x+2)=0

⇒ x−2=0 or x+2=0

⇒ x=2 or x=−2

∴ Zeroes of p(x)are 2 and -2.

Now,  sum of zeroes =2+(−2)=0=−01=−coefficient ofxcoefficient ofx2

and   product of zeroes =(2)(−2)=−4=−41=coefficient termcoefficient ofx2  Hence Verified.

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Answer 2

Given:

A quadratic polynomial x2 - 4.

To find:

The zeroes of the given quadratic polynomial. Also, verify the relationship between zeroes and coefficients.

Solution:

As given, we have a quadratic polynomial

${x}^{2} - 4$

This can be written as

${(x)}^{2} - \: {(2)}^{2}$

$= (x + 2)(x - 2)$

$[as \: {x }^{2} - {y}^{2} = (x - y)(x + y)]$

So, the value of the polynomial will be zero, if

$x = - 2 \: and \: 2$

Now,

as we know that in a polynomial ax2 + bx + c, the sum of zeroes is

$= \frac{ - b}{a}$

and

product of zeroes

$= \frac{c}{a}$

Where a, b and c are coefficients.

Now,

In the given polynomial x2 - 4, a = 1, b = 0 and c = -4.

So,

$\frac{ - b}{a} = \frac{0}{1} = 0$

$\frac{c}{a} = \frac{ - 4}{1} = - 4$

Also,

the sum of zeroes

$- 2 + 2 = 0$

Product of zeroes

$- 2 \times 2 = - 4$

As

$Sum \: of \: zeroes = \frac{ - b}{a} = 0$

and

$Product \: of \: zeroes = \frac{c}{a} = - 4$

Hence, the zeroes of the given quadratic polynomial are -2 and 2.

Also, the relationship between zeroes and coefficients is verified.